1. Check if the vector (1,2,3,4) belongs to the column space of the matrix
[ 1 2 3 ]
[ 4 5 6 ]
[ 7 8 9 ]
[ 0 1 2 ]

A vector that belongs to the column space of this matrix is a linear combination of the three columns. So, we have the following system of equations: 
 
					 x + 2y + 3z = 1 
					4x + 5y + 6z = 2 
					7x + 8y + 9z = 3 
					      y + 2z = 4.

This system has the following augmented matrix:

> A:= matrix([[1,2,3,1],[4,5,6,2],[7,8,9,3],[0,1,2,4]]);
A := [ 1 2 3 1 ]
[ 4 5 6 2 ]
[ 7 8 9 3 ]
[ 0 1 2 4 ]

> A:= gaussjord(A);
A := [ 1 0 -1 0 ]
[ 0 1 2 0 ]
[ 0 0 0 1 ]
[ 0 0 0 0 ]

This vector does not belong to the matrix's vector space since the Gauss-Jordan form of the augmented matrix has no solution. Since column three is impossible.

0*x + 0*y + 0*z = 1 is impossible.


2. Find the null space of the matrix
[ 1 2 3 4 5 ]
[ 2 3 4 5 6 ]
[ 1 0 2 0 6 ]

The kernel or null space of a matrix is the set of solutions of the homogeneous system of equations Av=0, where v = (x,y,z,t,s). The augmented matrix of this system is:
[ 1 2 3 4 5 0 ]
[ 2 3 4 5 6 0 ]
[ 1 0 2 0 6 0 ]

The Gauss-Jordan elimination procedure gives the following matrix:

> gaussjord(B);
[ 1 0 0 -4/3 0 0 ]
[ 0 1 0 5/3 -2 0 ]
[ 0 0 1 2/3 3 0 ]

Hence, the general solution is 
 
                                 x = (4/3)*t 
                                 y = (-5/3)*t + 2*s 
                                 z = (-5/3)*t + 2*s 
                                 t = t 
                                 s = s,

or in vector notation:

(x,y,z,t,s) = ((4/3)*t, (-5/3)*t + 2*s, (-5/3)*t + 2*s , t, s) = (4/3,-5/3,-5/3,1,0)t + (0,2,2,0,1)s.


So, the null space is spanned by the vectors (4/3,-5/3,-5/3,1,0) and (0,2,2,0,1).

3. Let p1 = x3 + x + 1, p2 = 2x3 - x2, p3 = x2 + x + 1, and q1 = x3 - 1, q2 = x3 - x2 + 1, q3 = x3 + x be polynomials. Is it true that

span(p1,p2,p3) =
span(q1,q2,q3) ?


If span(p1,p2,p3) = span(q1,q2,q3) then p1, p2, and p3 can be represented as linear combinations of q1, q2, and q3, and vice versa. This is not true. It is not possible to form p1 from a linear combination of q1, q2, and q3. We can prove this by attempting to solve the following equation.

p1 = a*q1 +b*q2 + c*q3 x3 + x + 1 = a(x3 - 1) + b(x3 - x2 + 1) + c(x3 + x)


This can be written as the following system of equations: 
 
				 a + b    = 1 
				   -b     = 0 
				        c = 1 
				-a + b    = 1

This is impossible. If b = 0, then equationa 1 and 4 have solutions for a, 1 and -1 respectively. So, span(p1,p2,p3) is not equal to span(q1,q2,q3).

4. Find the kernel of the linear transformation from R4 to R2 with the following standard matrix.

To find the kernel of this matrix we use the same method as we used in problem #2. The augmented matrix of the system Av = 0 (v = (x,y,z,t)) is
C := [ 1 2 3 4 0 ]
[ 2 1 3 4 0 ]

> gaussjord(C);
[ 1 0 1 4/3 0 ]
[ 0 1 1 4/3 0 ]

So, the general solution is 
 
                                   x = -z - (4/3)*t 
                                   y = -z - (4/3)*t 
                                   z = z 
                                   t = t,

or in vector notation:

(x,y,z,t) = (-z-(4/3)*t, -z-(4/3)*t, z, t) = (-1,-1,1,0)z + (-4/3,-4/3,0,1)t.


The kernel is spanned by (-1,-1,1,0) and (-4/3,-4/3,0,1).

5. Is it true that the range of the transformation in the previous problem coincides with R2?

The range of a transformation is spanned by its column vectors. So, it is spanned by the vectors (1,2), (2,1), (3,3), and (4,4). The vectors (1,0) and (0,1) span the R2. So, to prove that the range of the transformation in the previous problem coincides with R2, we need to prove that the span of the column vectors is equal to the span of the basic vectors, i and j.

It is clear that the column vectors are spanned by the basic vectors i and j. Thus, it is only necessary to prove that i and j are spanned by the column vectors. We need to prove the following two systems of equations 
 
               x1* [1]  +  y1* [2]  +  z1* [3]  +  t1* [4]  =    [1] 
                  [2]         [1]         [3]         [4]       [0] 
 
and  
 
               x2* [1]  +  y2* [2]  +  z2* [3]  +  t2* [4]  =    [0] 
                  [2]         [1]         [3]         [4]       [1]

have solutions. If they do, the column vectors span i and j.

We can combine the augmented matrices of these two systems to get

> A:= matrix([[1,2,3,4,1,0],[2,1,3,4,0,1]]);
A := [ 1 2 3 4 1 0 ]
[ 2 1 3 4 0 1 ]

> gaussjord(A);
[ 1 0 1 4/3 -1/3 2/3 ]
[ 0 1 1 4/3 2/3 -1/3 ]

Looking at the row-echelon form of the augmented matrix we can see that the two systems both have an infinite number of solutions. So, the basic vectors are indeed spanned by the column vectors and the range of the transformation in problem #4 coincides with R2.